$f(x, y) = \dfrac{y^2}{x}$ What is the partial derivative of $f$ with respect to $x$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2yx - y^2}{x^2}$ (Choice B) B $\dfrac{2y}{x}$ (Choice C) C $\dfrac{-y^2}{x^2}$ (Choice D) D $\dfrac{-2y}{x^2}$
Solution: Taking a partial derivative with respect to $x$ means treating $y$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial f}{\partial x} &= \dfrac{\partial}{\partial x} \left[ \dfrac{y^2}{{x}} \right] \\ \\ &= y^2 \left( \dfrac{-1}{{x^2}} \right) \\ \\ &= \dfrac{-y^2}{{x^2}} \end{aligned}$ In conclusion, $\dfrac{\partial f}{\partial x} = \dfrac{-y^2}{x^2}$